PUMaC 2017 · 个人决赛（A 组） · 第 3 题
PUMaC 2017 — Individual Finals (Division A) — Problem 3
题目详情
- Triangle ABC has incenter I . The line through I perpendicular to AI meets the circumcircle of ABC at points P and Q , where P and B are on the same side of AI . Let X be the point such that P X ‖ CI and QX ‖ BI . Show that P B, QC, and IX intersect at a common point. 1
解析
- Let D = P X ∩ BC, E = QX ∩ BC . Angle chasing, ( ) ∠ A ∠ C ∠ B ◦ ◦ ∠ IP D = ∠ QIC = ∠ AIC − ∠ AIQ = 180 − − − 90 = = ∠ IBD = ∠ QEC. 2 2 2 It follows that IP BD, IQCE, P DEQ are cyclic. Let Y = P B ∩ QC and let Z be the second intersection of ( IP BD ) , ( IQCE ). By the radical axis theorem, I, Z, X and I, Z, Y are collinear, so I, X, Y are collinear, thus P B, QC, IX are concurrent at Y . Problem written by Bill Huang 2